A 500 kg beam is free to pivot about its
lower end and is supported by a horizontal
cable at its upper end, as shown in the
figure. (figure has angle 35 degrees between beam and the wall)
Assuming the mass of the beam is
uniformly distributed, what is the tension in
Torque = r * F
The Attempt at a Solution
Using pivot point at the end of the beam, the torque is 0.
weight acting down on the beam is mg*l/2*sin(35). L is not givin not sure where to go from there.